Sd i-module2- rajesh sir

Dept. of CE, GCE Kannur Dr.RajeshKN
Design of Beams and
One-Way Slabs
for Flexure, Shear and Torsion
Dr. Rajesh K. N.
Assistant Professor in Civil Engineering
Govt. College of Engineering, Kannur
Design of Concrete Structures

2
(Analysis and design in Module II, III and IV should be based on Limit State
Method. Reinforcement detailing shall conform to SP34)
MODULE II (14 hours)
Design of beams with rectangular singly & doubly reinforced sections
and flanged sections for flexure at ultimate limit state as per IS 456.
Design of one way slabs.
Shear, torsion and bond: Analysis and design with and without shear
reinforcement at ultimate limit state as per IS 456,
deflection control, development length, splicing, curtailment, code
requirements.
Reinforcement detailing.

Basic Design Concepts
• The design problem is somewhat the reverse of the analysis problem
• A complete design of a beam involves considerations of:
• safety under the ultimate limit states in flexure, shear, torsion and
bond,
• serviceability limit states of deflection, crack-width, durability etc.

Requirements of flexural reinforcement
Concrete Cover
Clear cover is the distance measured from the exposed concrete
surface (without plaster and other finishes) to the nearest surface
of the reinforcing bar (Ref. Cl. 26.4 & Tables 16, 16A).
Exposure
Condition
Minimu
m Grade
Nominal
Cover (mm)
Allowance permitted
Mild M 20 20 Can be reduced by 5mm for main
rebars less than 12mm dia
Moderate M 25 30
Severe M 30 45 } Can be reduced by 5mm if concrete
} grade is M35 or higherVery severe M 35 50
Extreme M 40 75
Nominal cover requirements based on exposure conditions

5
Spacing of Reinforcing Bars (Cl. 26.3, Page 45)
The Code specifies minimum and maximum limits for the
spacing between parallel reinforcing bars in a layer.
The minimum limits are necessary to ensure that the concrete can
be placed easily in between and around the bars during the
placement of fresh concrete.
The maximum limits are specified for bars in tension for the
purpose of controlling crack-widths and improving bond.

6
Minimum and Maximum Areas of Flexural Reinforcement
Minimum Flexural Reinforcement in Beams (Cl. 26.5.1.1)
y
st
fbd
A 0.85)( min
=
In the case of flanged beams, the width of the web bw should
be considered in lieu of b.
• Minimum Flexural Reinforcement in Slabs
The minimum reinforcement (Ast)min in either direction
in slabs is (Cl. 26.5.2)
min
0.15 %
( )
0.12 %
of for Fe 250
of for Fe 415
g
st
g
A
A
A
⎧
= ⎨
⎩
Maximum Flexural Reinforcement in Beams (Cl. 26.5.1 b & 26.5.1.2)
Tension reinforcement (Ast) as well as compression
reinforcement (Asc) in beams shall be limited to 0.04 bD

7
Requirements for DEFLECTION Control
•The cross-sectional sizes of flexural members is mostly selected from
deflections control criteria under service loads.
•The Code (Cl. 23.2a) specifies a limit of span / 250 to the final
deflection due to all loads (including long-term effects of temperature,
creep and shrinkage).
•The explicit computation of maximum deflection can be rather
laborious
•Code recommends certain span/effective depth (l / d) ratios which
are expected to satisfy the requirements of deflection control
(Cl. 23.2.1)

8
Code Recommendations for Span/Effective Depth Ratios
(Cl. 23.2.1)
ctbasic kkdldl ××= )()( max
( )
⎪
⎩
⎪
⎨
⎧
=
spanscontinuousfor26
spanssupportedsimplyfor20
spanscantileverfor7
/ basicdl
The modification factors kt (which varies with pt and fst) and kc (which
varies with pc) are as given in Fig. 4 and Fig. 5 of the Code
where

9
Guidelines for selection of member sizes
Selection of flexural member sizes depends on:
serviceability criteria (control of deflections and crack-
widths)
requirements related to the placement of reinforcement
other structural, economic and architectural considerations

10
•In general, the recommended ratio of overall depth (D) to width (b) in
rectangular beam sections is in the range of 1.5 to 2.
•Often, architectural considerations dictate the sizes of beams. In the
case of building frames, the width of beams should, in general, be less
than or equal to the lateral dimension of the columns into which they
frame
•In design practice, the overall depths of beams are often fixed in
relation to their spans. Span to overall depth ratios of 10 to 16 are
generally found to be economical in the case of simply supported and
continuous beams.
•From practical considerations, it is desirable to limit the number of
different beam sizes in the same structure to a few standard modular
sizes, as this will greatly convenience the construction of formwork,
and permit reusability of forms
General Guidelines for Beam Sizes

11
•The design problem is to determine the cross-sectional dimensions of a
beam, and the area of tension steel Ast required to resist a known
factored moment Mu.
•Accordingly, the design equation for flexure is:
DESIGN OF SINGLY REINFORCED RECTANGULAR SECTIONS
uRu MM ≤ max,uu xx ≤with
( )0.87 0.416uR y st uM f A d x= −We know, x xu u≤ ,maxfor
2
0.870.416
0.87 1
0.362
0.87 1
y st
u y st
ck
y stu st
y
ck
f A
M f A d
d f b
f AM A
f
bd bd f bd
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
Rearranging,

⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
=≡
100
1
100
87.02
t
ck
yt
y
u p
f
fp
fR
bd
MThus lim,tt pp ≤for
,
lim 2
0
0
0
lim
.1498 for Fe 250
.1389 for Fe 415
.1338 for Fe 500
ck
u
ck
ck
f
M
R f
bd
f
⎧
⎪
= = ⎨
⎪
⎩
0 87
0 362
,lim,max
.
.
y stu
ck
f Ax
d f bd
=
•The minimum value of d corresponding to the limiting case pt =
pt,lim is obtained from

13
Fixing Dimensions of Rectangular Section
• Slabs:
• b is taken as 1000 mm and d is governed by the limiting l/d
ratios for deflection control
• a trial value of d may be assumed as approximately l/25 for
simply supported spans, l/32 for continuous spans and l/8 for
cantilevers
•Beams:
•The value of b may be suitably fixed and the value of d
corresponding to any R ≤ Rlim is given by:
Rb
M
d u
=
where Mu is the factored moment
2
cover)clear( tie
φ
−φ−−= Dd

14
Determining Area of Tension Steel
For Mu = MuR and xu < xu,max,
( )
( )
0.87 0.416
u
st reqd
y u
M
A
f d x
=
−
where xu is obtained by solving ( )0 362 0 416. .u ck u uM f bx d x= −
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
−⎟
⎠
⎞
⎜
⎝
⎛
=
d
x
d
x
bdfM uu
cku 416.01362.0 2
In terms of xu/d,
Alternative: Use of Design Aids
• The relationship between and pt are available in the
form of charts or tables for various combinations of fy and fck in SP : 16
2
uR M bd≡
• The designed section should be checked for strength (MuR>Mu)and
deflection control (l/d ratio), and suitably redesigned if found
inadequate

15
A rectangular reinforced concrete beam, under ‘moderate’ exposure
condition, is simply supported on two masonry walls 300 mm thick and
7m apart (centre-to-centre). The beam has to carry, in addition to its
own weight, a distributed live load of 20 kN/m and a dead load of
15 kN/m. Design the beam section for maximum moment at midspan.
Assume Fe 415 steel and M25 concrete.
Design Problem 1
Take clear cover as 30 mm. (T 16 for moderate exposure condition )
Assume a trial cross-section b = 350 mm, and D = 700 mm (span/10).
Let d = D – 50 = 650 mm.
Effective span (Cl. 22.2 of Code)∴
7.0
(7.0 0.3) 0.65 7.35
m (distance between supports)
(clear span )
l
d
⎧
= ⎨
− + = +⎩
Taking the lesser value, 7.0 ml =
1. Trial section and Effective span
A rectangular reinforced concrete beam, under ‘moderate’ exposure
condition, is simply supported on two masonry walls 300 mm thick and
7m apart (centre-to-centre). The beam has to carry, in addition to its
own weight, a distributed live load of 20 kN/m and a dead load of
15 kN/m. Design the beam section for maximum moment at midspan.
Assume Fe 415 steel and M25 concrete.
Design Problem 1

16
Self-weight 25 0.35 0.73
kN m m mDLwΔ = × × = 6.125 kN/m
15 0 6 125 21 125 kN m. . .DLw∴ = + = =LLw 20.0 kN/m (given)
Factored load (as per Code):∴
)(5.1 LLDLu www += =1.5 (21.125 + 20.0) = 61.69 kN/m
Factored Moment (maximum at midspan)
2 2
61 69 7 0
8 8
. .u
u
w l
M
×
= =
= 377.836kNm.
2. Loads and moment

17
3. Fixing up b, d and D
= 0.1389×25 = 3.472 MPa
With b = 350 mm, for a singly reinforced section, the minimum
required effective depth d is:
bR
M
d u
lim
min =
6
377 84 10
3 472 350
.
.
×
=
×
= 557.57 mm.
( )0 362 0 416,lim ,max ,max. .u ck u uM f bx d x= −
For Fe 415 steel, Mu,lim = 0.1389 fck bd2
,
2
im
lim
u lM
R
bd
=For M 25 concrete,
Adopt D = 650 mm.
Assuming 25φ bars, 8φ stirrups and clear cover of 30 mm,
d = 650 – 30 – 8 – 25/ 2 = 600 mm
2
0 362 1 0 416,max ,max
. .u u
ck
x x
f bd
d d
⎛ ⎞
= −⎜ ⎟
⎝ ⎠

18
4. Determining (Ast)reqd
Alternatively, using ‘design aids’
For = 3.00 MPa,
for M 25 concrete and Fe 415 steel,
6
2 2
377 84 10
350 600
.uM
R
bd
×
= =
×
[ ]
0.87
0.87 0.416 0.87 0.416
0.362
0.87
y st
u y st u y st
ck
y st
u y st
ck
f A
M f A d x f A d
f b
f A
M f A d
f b
⎡ ⎤
= − = −⎢ ⎥
⎣ ⎦
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
Solving this quadratic equation 2090 2
mm( )st reqdA =
2090 2
mm( )st reqdA =pt = 0.997 [Table 3, Page 49, SP: 16],
6 415
377.84 10 0.87 415 600
25 350
st
st
A
A
⎡ ⎤
× = × × −⎢ ⎥×⎣ ⎦

5. Detailing
Provide 5 # 25 ϕ bars
Ast, provided = 2454.37 mm2
650
600
350
5 – 25φ
30+8+25/2
6. Design Checks
6.(a) Check for strength in flexure
Actual d = 650 – 30 – 8 – 25/2 = 600 mm
MuR = 428.5kNm > Mu = 377.84 kNm — Hence, safe.
0.87
y st
uR y st
ck
f A
M f A d
f b
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
pt, provided 100 1.1687stA
bd
= × =
415 2454.37
0.87 415 2454.37 600
25 350
×⎡ ⎤
= × × −⎢ ⎥×⎣ ⎦

6.(b) Check for deflection control:
For =1.1687, andpt
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
)(
)(
58.0
providedst
reqst
ys
A
A
ff
2090
0 58 415
2454 4
.
.
⎛ ⎞
= × ×⎜ ⎟
⎝ ⎠
= 204.96 MPa
kt = 1.1 (from Fig. 4 of Code),
and, as pc = 0 (singly reinforced beam), kc = 1 (from Fig. 5 of Code).
20 1 1 1
max
.
l
d
⎛ ⎞
= × ×⎜ ⎟
⎝ ⎠
= 22
7000
600provided
l
d
⎛ ⎞
=⎜ ⎟
⎝ ⎠
= 11.6 <
max
l
d
⎛ ⎞
⎜ ⎟
⎝ ⎠
— Hence, OK.

21
1. Select a trial section and get effective span
Design Procedure summarised (Singly reinforced rectangular sections)
2. Find loads and factored moment
distance between supports
clear span
l
d
⎧
= ⎨
+⎩
Take the lesser value.
3. Fix up b, d and D
lim
min
uM
d
R b
∴ =
E.g., For Fe 415 steel, Mu,lim = Rlim bd2 = 0.1389 fck bd2
Adopt a suitable D, thus get d assuming
diameter of main bars, stirrups and clear
cover.

22
4. Determining (Ast)reqd
Alternatively, use SP:16 to get pt for a given value of and for
the given grades of steel and concrete.
2
uM
bd
0.87
y st
u y st
ck
f A
M f A d
f b
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
Solving this quadratic
equation, get ( )st reqdA
5. Design Checks
5.(a) Check for strength in flexure
Find actual d provided and get
Ensure MuR > Mu . Else, redesign (e.g., by increasing d).
0.87
y st
uR y st
ck
f A
M f A d
f b
⎡ ⎤
= −⎢ ⎥
⎣ ⎦

5.(b) Check for deflection control:
For provided and,t cp p
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
)(
)(
58.0
providedst
reqst
ys
A
A
ff
Get kt and kc from Fig. 4 & 5 of Code
max
l
d
⎛ ⎞
⎜ ⎟
⎝ ⎠ maxprovided
l l
d d
⎛ ⎞ ⎛ ⎞
<⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
and ensureGet Else, redesign.
6. Detailing

Design Procedure for Given Mu
• For a given rectangular section the Mu,lim for a singly reinforced
section should be first determined
• If Mu,lim is greater than or equal to the factored moment Mu, the
section should be designed as a singly reinforced section
• Otherwise (for Mu > Mu,lim), the section should be designed as a
doubly reinforced section
Design of doubly reinforced rectangular sections

ΔMu is obtained from a couple comprising a (compressive) force (fsc –
0.447fck)Asc and an equal and opposite (tensile) force (0.87fy ∆ Ast ),
with a lever arm ( )d d′−
ΔA
M M
f d d
st
u u
y
=
−
− ′
,
. ( )
lim
0 87
A
f A
f f
sc
y st
sc ck
=
−
( . )( )
.447
087
0
Δ

Design Formula M M M Mu uR u u= = +,lim Δ
• where Mu,lim is the limiting moment capacity of a singly
reinforced ‘balanced’ section
• ΔMu is the additional moment capacity desired from the
compression steel Asc and the corresponding additional tension steel
• ∆ Ast is the additional tension steel corresponding to the
compression steel Asc
p p pt t lim t= +, ΔA A Ast st st= +,lim Δ
Ast,lim is the tension steel corresponding to Mu,lim
∆ Ast is the additional tension steel corresponding to ΔMu

Design the flexural reinforcement for the beam in Design Problem 1,
if Mu = 520 kNm
Design Problem 2
Given b=350 mm and D = 650 mm.
Since the breadth and depth of the section are fixed here, it is
necessary to check whether the section is singly or doubly reinforced.
Effective span l = 7.0 m
Mu,lim = 0.1389 fck bd2 = 437.54 kNm < Mu
Hence the section has to be doubly reinforced.
For Fe 415 steel,
1. Singly reinforced/doubly reinforced?
Let d = D–50 =600 mm.

28
0 362
0 87
,max
,lim
.
.
ck u
st
y
f bx
A
f
=
2. Determining Ast,lim
( )0 87 0 416
,lim
,lim
,max. .
u
st
y u
M
OR A
f d x
=
− ×
2
2526 6 mm,lim .stA =
3. Determining ΔAst
Assuming 20 mm φ bars for compression steel,
mm48≈′d (30 mm clear cover + 8 mm stirrup + φ/2 )
( )
( )0 87
lim,
.
u u
st reqd
y
M M
A
f d d
−
Δ =
′−
= 413.75 mm2
( )
( )
6
520 437 54 10
0 87 415 600 48
.
.
− ×
=
× × −
( ), ,limst reqd st st reqd
A A A= + Δ = 2526.6 + 413.75 = 2940.35 mm2
Provide 4 nos 32 mm ϕ [Ast,provided =3216.9mm2]
4. Determining Ast

Actual d (assuming 30 mm clear cover and 8 mm stirrups):
d = 650-(30+8+32/2) = 596 mm < 600 mm assumed earlier
Revising the above calculations with d = 646 mm, assuming 20mm φ
compression steel,
Mu,lim = 0.1389 fck bd2 = 431.7 kNm,
( )0 87
lim,
,
.
u u
st reqd
y
M M
A
f d d
−
Δ =
′−
,st reqdA
= 446.18 mm2
= 2509.7 + 446.18 = 2955.96 mm2,
Hence actual (ΔAst) provided = 3216.9 – 2509.7 =707.2 mm2
( )
( )
6
520 431 7 10
0 87 596 48
.
. yf
− ×
=
−
Ast,provided = 3216.9mm2
( )
0
0 87 0 416
2
25 9 7 mm,lim
,lim
,max
.
. .
u
st
y u
M
A
f d x
= =
− ×
5. Revising steel areas based on the actual d provided

( )0 87
0 447
,
.
.
y st
sc reqd
sc ck
f A
A
f f
Δ
=
−
( )
( )
0 87 415 707 2
361 0 447 25
. .
.
× ×
=
− ×
= 729.9 mm2
Provide 4 nos 16 mm ϕ
[Asc = 804. 2 mm2 > 729.9 mm2 ].
6. Determining Asc
48
0 0035 1 0 0035 1
0 48 596,max
. .
.
sc
u
d
x
ε
⎛ ⎞′ ⎛ ⎞
= × − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠
= 0.002913
0 87 361 MPa.sc yf f≅ =From design stress-strain curve
350
650
30 clear
30 clear
54
4 – 16 φ
4 – 32 φ
beam section
46
8 φ

7. Design check
Actual d provided: d = 596 mm 30 16 2 8 46d mm′ = + + =
46
0 0035 1
0 48 596
.
.
scε
⎛ ⎞
= −⎜ ⎟
×⎝ ⎠
= 0.00294
( )0 87
0 447
lim,*
.
.
y st st
sc
sc ck
f A A
A
f f
−
=
−
( )0 87 415 3216 9 2509 7
361 0 447 25
. . .
.
× × −
=
− ×
= 729.9 mm2
*
,sc provided scA A> Section is under-reinforced.
0 87 361 MPa.sc yf f≅ =From design stress-strain curve
Hence, OK.
Asc,provided = 804. 2 mm2
(Note: Else, if , increase Asc )
*
,sc provided scA A<

8. Check for deflection control
3216 9
100 1 542
350 596
.
.tp = × =
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
)(
)(
58.0
providedst
reqst
ys
A
A
ff
2955 96
0 58 415
3216 9
.
.
.
⎛ ⎞
= × ×⎜ ⎟
⎝ ⎠
= 221.18 MPa
kc = 1.15 (from Fig. 5 of Code).
20 1 0 1 15
max
. .
l
d
⎛ ⎞
= × ×⎜ ⎟
⎝ ⎠
= 23
7000
596provided
l
d
⎛ ⎞
=⎜ ⎟
⎝ ⎠
= 11.75 <
max
l
d
⎛ ⎞
⎜ ⎟
⎝ ⎠
— Hence, OK.
804 2
100 0 3855
350 596
.
.cp = × =
×

1. For the given b and D, and effective span l check whether the
section is singly or doubly reinforced.
E.g., For Fe 415 steel, Mu,lim = 0.1389 fck bd2
Design Procedure summarised (Doubly reinforced rectangular sections)
If Mu,lim < Mu the section has to be doubly reinforced.
0 362
0 87
,max
,lim
.
.
ck u
st
y
f bx
A
f
=
2. Determine Ast,lim
( )0 87 0 416
,lim
,lim
,max. .
u
st
y u
M
OR A
f d x
=
− ×
3. Determine ΔAst Assuming dia of compression steel,
( )
( )0 87
lim,
.
u u
st reqd
y
M M
A
f d d
−
Δ =
′−

Get actual d and revise the calculations of Mu,lim = 0.1389 fck bd2
( )0 87
lim,
,
.
u u
st reqd
y
M M
A
f d d
−
Δ =
′−( )0 87 0 416
,lim
,lim
,max. .
u
st
y u
M
A
f d x
=
− ×
5. Revise steel areas based on the actual d provided
lim, , ,st reqd st st reqdA A A= + Δ
Hence actual lim, , ,st provided st provided stA A AΔ = −
( ), ,limst reqd st st reqd
A A A= + Δ
Choose dia and number of bars. Get Ast,provided
4. Determine Ast

( )0 87
0 447
,
.
.
y st
sc reqd
sc ck
f A
A
f f
Δ
=
−
6. Determine Asc
0 0035 1
,max
.sc
u
d
x
ε
⎛ ⎞′
= × −⎜ ⎟⎜ ⎟
⎝ ⎠
scfFrom design stress-strain curve find corresponding to
Choose dia and number of bars.
Get Asc,provided
7. Design check
Get actual d provided.
46
0 0035 1
0 48 596
.
.
scε
⎛ ⎞
= −⎜ ⎟
×⎝ ⎠
( )0 87
0 447
lim,*
.
.
y st st
sc
sc ck
f A A
A
f f
−
=
−
scfFrom design stress-strain curve find corresponding to
If *
,sc provided scA A> section is under-reinforced.
Else, if , increase Asc
*
,sc provided scA A<

For provided and,t cp p
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
)(
)(
58.0
providedst
reqst
ys
A
A
ff
Get kt and kc from Fig. 4 & 5 of Code
max
l
d
⎛ ⎞
⎜ ⎟
⎝ ⎠ maxprovided
l l
d d
⎛ ⎞ ⎛ ⎞
<⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
and ensureGet Else, redesign.
9. Detailing

ONE-WAY SLABS AND TWO-WAY SLABS
• When the slab is supported only on two opposite sides, the slab
bends in one direction only. Hence, it is called a one-way slab.
• When the slab is supported on all four sides, and the length is less
than twice the width, it bends in two directions. Hence, it is called a
two-way slab.
one–way
slab two–way slab
l < 2 b

b
l > 2 b
one–way
slab
• If the length is greater than about twice the width, the bending
along the longitudinal direction is negligible compared to the
transverse (short-span) direction, and the resulting slab action is one-
way.

• To resist secondary moments (arising as a result of the necessity to
resist lateral deformation).
• To resist bending moments in the transverse direction generated
locally when the slab is subject to concentrated loads.
• To resist secondary stresses introduced by shrinkage and
temperature effects
Distribution steel

Design a one-way slab, with a clear span of 5.0m, simply supported on
230 mm thick masonry walls, and subjected to a live load of 3 kN/m2
and a surface finish load of 1 kN/m2, using Fe 415 steel and M25
concrete. Assume that the slab is subjected to mild exposure condition.
1. Trial depth and effective span
Assume an effective depth
5000
25
d ≈ = 200 mm
Problem 3
Let overall depth 200 40 240 mmD = + =
Effective span
5000 230 5230
5000 200 5200
mm (c/c distance)
mm
l
+ =⎧
= ⎨
+ =⎩
Taking the lesser values (as per Code), l = 5.2 m.
Design of one-way slabs

Distributed load due to self-weight,
25 0 24 6 03 2
kN m m kN m. .DLwΔ = × =
6 0 1 0 7 0 2
kN m. . .DLw = + = 3 0 2
kN m.LLw =
Factored load )(5.1 LLDLu www += = 1.5 (7.0+3.0) =15.0 kN/m2
Factored Moment (maximum at midspan)
2
8
u
u
w l
M = 50 7kNm.=
2
15 0 5 2
8
. .×
=
2. Loads and moment (per metre width)

3. Determining Ast (main bars)
0.87
y st
u y st
ck
f A
M f A d
f b
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
6 415
50.7 10 0.87 415 200
25 1000
st
st
A
A
⎡ ⎤
× = × × −⎢ ⎥×⎣ ⎦
Alternatively, using ‘design aids’
For = 1.2675 MPa,
for M 25 concrete and Fe 415 steel,
6
2 23
50 7 10
20010
.uM
R
bd
×
= =
×
Solving this quadratic equation 749 2
mm( )st reqdA =
[Table 3, Page 49, SP: 16],
( )
( )
( )
0 385 0 369
0 369 1 2675 1 25
1 3 1 25
. .
. . .
. .
tp
−
= + × −
−
0 375
1000 200 749 2
100
2
mm
.
( ) .st reqdA = × × =
0 375.=

Spacing of bars
1000
st
A
s
A
φ
=
Provide 12 mm ϕ bars.
2
1000 12
4
749
π⎛ ⎞
× ×⎜ ⎟
⎝ ⎠= 151mm=
3 3 200 600
300 mm
v
d
s
= × =⎧
≤ ⎨
⎩
Maximum spacing (Cl.26.3.3 a)
Provide 10 mm ϕ bars at 150 mm c/c spacing as main reinforcement
4. Distribution bars (Cl. 26.5.2.1, Page 48)
0 0012 1000 240( ) .st distA = × × = 288 mm2/m.
Provide 8 mm ϕ distribution bars.
Spacing
1000
st
A
s
A
φ
=
2
1000 8
4
288
π⎛ ⎞
× ×⎜ ⎟
⎝ ⎠= 174 53mm.=

5 5 200 1000
450 mm
v
d
s
= × =⎧
≤ ⎨
⎩
Maximum spacing (Cl.26.3.3 b)
Provide 8 mm ϕ bars at 170 mm c/c spacing as distribution reinforcement
5. Detailing
• The complete detailing of the slab is given in Figure. This meets the
Code requirements.
• Alternate bars of the main reinforcement are bent up (cranked) near
the supports at a distance of 0.1 l from the support (Cl. D-1.6, Page 90)
• This is in order to resist any flexural tension that may arise due to
partial fixity at the support

SECTION A– A
12φ @ 300c/c distributors 8 φ
230
520 12φ @ 150 c/c
(alternate bars
bent up)
240 mm
clear cover 30 mm8 φ@ 170 c/c
(distributors)
8φ @ 170 c/c
12φ @ 150c/c
520
240 mm THICK
M25
Fe415
PLAN
(0.1L)
A A
230 2305000
Details of a one-way slab

490 87
100
1000 205,
.
t providedp = ×
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
)(
)(
58.0
providedst
reqst
ys
A
A
ff
478 2
0 58 415
490 87
.
.
.
⎛ ⎞
= × ×⎜ ⎟
⎝ ⎠
= 234.487 MPa
20 1 7
max
.
l
d
⎛ ⎞
= ×⎜ ⎟
⎝ ⎠
= 34
4200
205provided
l
d
⎛ ⎞
=⎜ ⎟
⎝ ⎠
= 20.4<
max
l
d
⎛ ⎞
⎜ ⎟
⎝ ⎠
— Hence, OK.
Providing a clear cover of 30 mm, d = 240–30–10/2 = 205 mm.
2
1000 10
4
160
,st providedA
π⎛ ⎞
× ×⎜ ⎟
⎝ ⎠=
490 87 2
mm.= 0 2395.=

47
Curtailment of flexural tension reinforcement
• For economy, it is desirable to progressively curtail (‘cut-off’) the
flexural tension reinforcement, according to the decrease in bending
moment.
• Other factors to be considered in fixing actual bar cut-off points
• unexpected shifts in maximum moments,
• development length requirements,
• influence on shear strength
• development of diagonal tension cracks due to the effects of
discontinuity.
• Accordingly, Code (Cl. 26.2.3, Page 44) specifies a number of
requirements

Theoretical Bar Cut-off Points
• The ‘theoretical cut-off point’ for a bar is that point beyond which it
is (theoretically) no longer needed to resist the design moment.
• In a prismatic beam the required area of tension reinforcement
varies nearly linearly with the bending moment
• The stress at the end of a bar is zero; it builds up gradually along its
length through bond with the surrounding concrete.
• In order to develop the full design stress (0.87fy) in the bar at a
section, a minimum development length Ld is required on either side of
the section. (Ref: Cl. 26.2.1 of Code)
4
s
d
bd
L
φσ
τ
= s
bd
φ
σ
τ
=
=
=
Nominal diameter of the bar
Stress in bar at the section considered
Design bond stress given in Cl. 26.2.1.1
Development length requirements:

Code Requirements
1.The shear at the cut-off point does not exceed two-thirds of the
shear resisting capacity of the section.
2.Excess stirrups are provided over a distance of 0.75d from the cut-
off point having an area Asv and a spacing sv such that:
A
b s
f
sv
w v
y
≥
0 4.
s
d
v
b
≤
8β
is the ratio of the area of bars cut off to the total area of the bars
at the section.
βb
3. For 36 mm ϕ and smaller bars,
1. the continuing bars provide at least twice the area required for
flexure at the cut-off point and
2. the shear does not exceed three-fourth of the shear resisting
capacity of the section.

• The extension of bars beyond the theoretical cut-off points should
not be less than the effective depth or 12 times the bar diameter
• If the anchorage length beyond the zero moment location is
denoted as Lo , then as per Code requirement,
( )M V L LuR u o d
*
+ ≥
• where (MuR/Vu)* = 1.3MuR/Vu at a simple support with a confining
reaction, and MuR/Vu otherwise.
• At a point of inflection, the anchorage length Lo is limited to the
effective depth d or 12φ, whichever is greater.

Splicing of reinforcement (Cl. 26.2.5)

Critical sections for shear design
Design for SHEAR
• The maximum shear force usually occurs in a flexural member at
the face of the support,
• When a support reaction introduces transverse compression in the
end region of the member, critical section is located at a distance d
(effective depth) from the face of the support (Cl. 22.6.2.1, Page 36)

1. stirrups perpendicular to the beam axis;
2. stirrups inclined (at 45° or more) to the beam axis; and
3. longitudinal bars bent-up (usually, not more than two at a time) at
45° to 60° to the beam axis, combined with stirrups.
Types of Shear Reinforcement

Design shear strength without shear reinforcement
uc cV bdτ= where is obtained from Table19, Page 73cτ
Design shear strength with shear reinforcement (Cl. 40.1, Page 72)
For beams of uniform depth, nominal shear stress at a section is
defined as:
u
v
V
bd
τ =
where is the factored shear force at the sectionuV

Code Recommendations on shear reinforcement
If is less than (Table19, Page 73) minimum stirrup shear
reinforcement has to be provided
vτ cτ
If is between suitable shear reinforcement has
to be designed for shear (as per Cl. 40.4, Page 72)
vτ and ,maxc cτ τ
If exceeds (Table 20, Page 73) the section has to be
suitably redesigned — by increasing the cross-sectional area and/or
improving the grade of concrete
vτ τ c,max
Minimum stirrup reinforcement (Cl. 26.5.1.6, Page 48)
0 4
0 87
.
.
sv
v y
A
bs f
≥
Minimum shear reinforcement in the form of stirrups should be
provided so that:
0 5.v cτ τ>for

Design of shear reinforcement (Cl. 40.4)
us u uc u cV V V V bdτ= − = −
Shear reinforcement has to be provided for a shear force of:
Area of shear reinforcement required is calculated from:
0 87. y sv
us
v
f A d
V
s
= for vertical stirrups
( )
0 87.
sin cosy sv
us
v
f A d
V
s
α α= + for inclined stirrups or
bars bent up at different cross sections
0 87. sinus y svV f A α= for bars bent up at same cross section

• The Code limits the value of the spacing sv to 0.75d for ‘vertical’
stirrups, and d for inclined stirrups with α = 45°
• In no case shall the spacing exceed 300 mm
Spacing of shear reinforcement (Cl. 26.5.1.5, Page 47)

Check the adequacy of the given section with regard to shear. Assume
that a support reaction introduces transverse compression in the end
region of the member. End supports are 300 mm wide. Take clear span
= 6.7 m and
Problem 3
650
600
350
3 – 25φ
30+8+25/2
uw (factored load) = 61.69 kN/m
Max. factored shear force (at d from support)
( )
61 69 7
61 69 0 15 0 6
2
.
. . .uV
×
= − × +
169 64 kN.=
Nominal shear stress at the section is:
u
v
V
bd
τ =
3
169 64 10
350 600
. ×
=
×
0 807MPa.= ( )3 1MPa,max .cτ< =
(Table 20, Page 73)

Ast, provided = 1472.62 mm2 100
0 7.sA
bd
=
(Table 19, Page 73)
( )
( )
( )
0 57 0 49
0 49 0 7 0 50
0 75 0 50
. .
. . .
. .
cτ
−
= + × −
−
0 554. vτ= <
Since is between suitable shear reinforcement
has to be designed (as per Cl. 40.4, Page 72)
vτ and ,maxc cτ τ

169 64 116 34 53 3 Nk. . .us u ucV V V= − = − =
0 87. y sv
us
v
f A d
V
s
=
0 75 0 75 600 450
300 mm
. .
v
d
s
= × =⎧
≤ ⎨
⎩
Hence provide 300 mm spacing (least of the above).
Maximum spacing
To find shear reinforcement)
Shear reinforcement has to be provided for a shear force of:
Find out spacing of 2–legged 8ϕ vertical stirrups from
2
0 87 415 2 8 6000 87 4 409
53 3
mm
..
.
y sv
v
us
f A d
s
V
π
× × × × ×
= = =
Assume 2–legged 8ϕ vertical stirrups are provided.

Total design shear strength 116 34 149 93. .= +
0 75 0 75 600 450
300 mm
. .
v
d
s
= × =⎧
≤ ⎨
⎩
Spacing of stirrups (Cl. 26.5.1.5, Page 47)
Hence safe.
Hence shear reinforcement = 2–legged 8ϕ vertical stirrups at 300 mm spacing.
Maximum allowed spacing
( )266 27 233 27kN. .uV>= =
Design shear strength without shear reinforcement
uc cV bdτ= 0 554 350 600.= × × 3
116 34 10 N.= ×
Design shear strength with a shear reinforcement of 2–legged 8ϕ stirrups
=100.53 mm2
2
2 8
4
,sv providedA
π
= × ×
0 87. sv
us
v
A d
V
s
=
0 87 100 53 600
350
. .× ×
= 149 93kN.=
Provide 2–legged 8ϕ vertical stirrups at 350 mm spacing.
Alternate calculation (to find shear reinforcement)

Design for TORSION
Torsional reinforcement has to be suitably designed when the torsional
shear stress exceeds the shear strength of the sectionτ t τ c
Where flexural shear Vu occurs in combination with torsional shear (as
is commonly the case), the combined shear stress has to be considered.
The equivalent nominal shear stress (Cl. 41.3.1, Page 75)
1 6.u u
ve
V T b
bd
τ
+
=
If exceeds (Table 20, Page 73) the section has to be
suitably redesigned — by increasing the cross-sectional area (especially
width) and/or improving the grade of concrete
veτ τ c,max

If is less than minimum stirrup reinforcement has to be
provided
veτ cτ
If is between suitable torsional reinforcement
(both transverse and longitudinal) has to be designed for the combined
effects of shear and torsion (as per Cl. 41.4)
veτ and ,maxc cτ τ

Code Provisions for Transverse Reinforcement
If Tu and Vu are the factored twisting moment and factored shear
force at the section under consideration,
A
T s
b d f
V s
d f
sv
u v
y
u v
y
= +
1 1 1087 2 5 087( . ) . ( . )
Asv = 2At is the total area of two legs of the stirrup;
sv is the centre-to-centre spacing of the stirrups;
b1 and d1 are the centre-to-centre distances between the corner bars
along the width and depth respectively
The Code (Cl. 41.4.3) specifies a minimum limit to the total area of
transverse reinforcement:
A
bs
f
sv
ve c v
y
≥
−( )
.
τ τ
0 87

• The torsional moment Tu is converted into an effective bending
moment Mt defined as:
Code Provisions for Longitudinal Reinforcement
( )1
1 7.
t u
D b
M T
+
=
• ‘Equivalent bending moments’
M M Me t u1 = + M M Me t u2 = −
• The longitudinal reinforcement area Ast is designed to resist the
equivalent moment Me1, and this steel is to be located in the ‘flexural
tension zone’.
• In addition, if Me2 > 0 (i.e., Mt > Mu), then a reinforcement area Ast’ is
to be designed to resist this equivalent moment, and this steel is to be
located in the ‘flexural compression zone’.

Check the adequacy of the given section with regard to shear and
torsion, assuming a torsional moment per metre run of 5kNm. Assume
fixity at the ends of the beam against torsion and flexure. Take effective
span = 7 m and
Problem 4
650
600
350
5 – 25φ
30+8+25/2
uw (factored load) = 61.69 kN/m
Max. twisting moment (at support)
Max. bending moment (at support)
Max. shear force (at support)
T
t l
u
u
=
2
2
u
u
w l
V =
5 7
17 5
2
kNm.
×
= =
2
12
u
u
w l
M =
2
61 69 7
12
. ×
=
61 69 7
2
. ×
=
251 9 kNm.=
215 915 kN.=
Stress resultants

The equivalent nominal shear stress
1 6.u u
ve
V T b
bd
τ
+
=
3 6
215 915 10 1 6 17 5 10 350
350 600
. . .× + × ×
=
×
1 409MPa.=
(Table 19, Page 73)
( )
( )
( )
0 70 0 64
0 64 1 168 1 0
1 25 1 0
. .
. . .
. .
cτ
−
= + × −
−
Ast, provided = 2454.37 mm2 100
1 168.sA
bd
=
0 68. veτ= <
3 1MPa,max .cτ = (Table 20, Page 73) ,maxc veτ τ>
Stresses
Since is between suitable torsional reinforcement
(both transverse and longitudinal) has to be designed for the combined
effects of shear and torsion (as per Cl. 41.4)
veτ and ,maxc cτ τ

Check for adequacy of longitudinal reinforcement
( )1
1 7.
t u
D b
M T
+
=
( )1 650 350
17 5
1 7
.
.
+
= 29 4kNm.=
1e t uM M M= +
2 0e t uM M M= − <
29 4 251 9. .= + 281 3kNm.=
Ultimate resisting moment of section
Ast = 2454.37 mm2 (pt = 1.168) at bottom ‘flexural tension zone’
0.87
y st
uR y st
ck
f A
M f A d
f b
⎡ ⎤
= −⎢ ⎥
⎣ ⎦
415 2454.37
0.87 415 2454.37 600
25 350
×⎡ ⎤
= × × −⎢ ⎥×⎣ ⎦
= 428.5kNm 281 3kNm.eM> =
Hence longitudinal reinforcement is sufficient.

Check for adequacy of transverse reinforcement
Area of 2–legged 8ϕ stirrups provided, Asv = 100.53 mm2
This should exceed the requirements given by Cl 41.4.3
A
T s
b d f
V s
d f
sv
u v
y
u v
y
= +
1 1 1087 2 5 087( . ) . ( . )
sv = 150 mm (say), b1 = 350-2x50=250 mm, d1 = 650-50-44=556 mm
( ) ( )
3 3
17 5 10 150 215 915 10 150
250 556 0 87 415 2 5 556 0 87 415
. .
. . .
svA
× × × ×
= +
× × × ×
= 64.59 mm2
< Asv provided

( )
0 87.
ve c v
sv
y
bs
A
f
τ τ−
=
Minimum limit to the total area of transverse reinforcement (Cl. 41.4.3):
( )1 409 0 68 350 150
0 87 415
. .
.
− × ×
=
×
( ) ( )
1
1 1
250 25 10 283
4 283 582 5 4 216 38
300 mm
. .v
x
s x y
= + + =⎧
⎪
≤ + = + =⎨
⎪
⎩
sv provided = 150 mm < sv required. Hence safe.
= 106 mm2
> Asv provided
Spacing of torsion reinforcement (Cl. 26.5.1.7, Page 48)
Hence provide 2–legged 10 ϕ stirrups, Asv = 157 mm2

Summary
Design of beams with rectangular singly & doubly reinforced sections
and flanged sections for flexure at ultimate limit state as per IS 456.
Design of one way slabs.
Shear, torsion and bond: Analysis and design with and without shear
reinforcement at ultimate limit state as per IS 456,
deflection control, development length, splicing, curtailment, code
requirements.
Reinforcement detailing.

Sd i-module2- rajesh sir

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